What is the Chemical Potential (μ)?

If you're reading this, you've probably encountered the concept of the chemical potential. You might even be able to state that it's equal to the partial molar Gibbs energy. But this formal definition can obscure why chemical potential is such a central quantity when considering chemical reactions. For comparison, when discussing temperature, we rarely begin by defining it as the partial derivative of internal energy with respect to entropy. Instead, we focus on its role in determining the direction of heat flow.

Flows of Heat and Mass

Consider two blocks at different temperatures brought into thermal contact. Energy flows as heat from the hot block to the cold one. The underlying reason for this process is that it increases the total entropy of the system. This entropy increase is what determines the direction of spontaneous change.

A similar picture holds for matter. Suppose we have two connected boxes, one containing a high concentration of a substance and the other containing a lower concentration. In that case, the material will diffuse from the high concentration to the low concentration. Again, the driving force is the increase in entropy for the system. We often describe this as diffusion from high to low concentration, but the deeper reason for the change is the resulting increase in entropy.

Now, let's say that these two boxes contain different solvents: water in the left box and oil in the right. Initially, only the water contains dissolved oxygen. What happens when the boxes are connected? At equilibrium, oxygen will be found in both boxes, but not necessarily in equal concentrations. Because oxygen is more soluble in oil than in water, its concentration will be higher in the oil phase at equilibrium. In this case, concentration alone doesn't determine the outcome; the solvation environment matters too. We say that oxygen is partitioned between the two solvents.

To describe both the effects of concentration and the energetic environment of a molecule, we introduce the chemical potential. The chemical potential is given the symbol $\mu$ and has units of J mol^-1. Just as temperature determines the direction of heat flow, chemical potential determines the direction of mass flow or, as will be seen below, the direction of a reaction. Particles tend to move from regions of higher to lower chemical potential, and equilibrium corresponds to the point where chemical potentials are equal. See here for a proof that at equilibrium the chemical potentials will be equal.

An important point about the chemical potential is that it is an intensive property; its value does not depend on the amount of material present. For example, if you have a 1 kg block of material at 30$^{\circ}$C, doubling the mass does not change the temperature.^[1] Temperature, like chemical potential, is a local property and remains unchanged when the system is scaled up. This contrasts with extensive properties such as the total Gibbs energy, which does depend on the size of the system. However, if we express these extensive quantities on a per mole basis, as is the case with the molar Gibbs energy, molar enthalpy, or molar entropy, we obtain intensive quantities. This is analogous to molar volume; the total volume is extensive, but dividing by the number of moles gives the intensive molar volume.

The Gibbs Energy and Entropy

The Gibbs energy is often used in chemistry as it is the natural thermodynamic variable to use when studying systems at constant temperature and pressure. You may have encountered Gibbs energy as a measure of the “useful work” a system can do under reversible conditions. But in the present context, it's more beneficial to think about its relationship to entropy. The usual form: $$ \Delta G = \Delta H - T\Delta S $$ can be reformulated. At constant pressure and temperature, the enthalpy change ($\Delta H$) corresponds to the heat exchanged with the surroundings, and so: $$ \Delta H = - T\Delta S_{\text{surroundings}} $$ Substituting this gives: $$ \begin{align*} \Delta G &= -T \Delta S_\mathrm{surroundings}- T\Delta S_\mathrm{system} \\ &= -T\Delta S_{\mathrm{universe}} \end{align*} $$ Consequently, under these conditions, the Gibbs energy change tells us the total entropy change of the universe. A negative $ \Delta G $ means the process increases total entropy, that is, it is spontaneous.

Gibbs Energy and Chemical Reactions

The aim of this article is to consider a simple gas-phase reaction and explain how the chemical potential drives changes in chemical reactions. To keep things straightforward, we will consider two ideal gases that interchange via the process A ⇌ B. This could be an example of an isomerisation, and assuming the gases both behave ideally then the molar Gibbs energy, $G_{\mathrm{m},i}$, for the ith species is given by: $$ G_{\mathrm{m},i} = G^{\minuso}_{\mathrm{m},i} + RT\ln \frac{p_i}{p^{\minuso}} $$ here $G^{\minuso}_{\mathrm{m},i}$ is the standard molar Gibbs energy for the pure species, R is the gas constant (8.314 J K^-1 mol^-1), T is the temperature (K), $p_i$ is the partial pressure of the gas and $p^{\minuso}$ is the standard pressure (1 bar).

For this reaction we will consider the total number of moles of A and B ($n_A$ and $n_B$ respectively), tracking their interconversion. The total Gibbs energy for the system is then, for these ideal gases, the sum of the two Gibbs energies: $$ G = n_A G_{\mathrm{m},A} + n_B G_{\mathrm{m},B} $$ expanding this out we have: $$ \begin{align*} G =~&n_A G^{\minuso}_{\mathrm{m},A} + n_B G^{\minuso}_{\mathrm{m},B} \\ &+ n_A RT \ln\left(\frac{p_A}{p^{\minuso}}\right) \\ &+ n_B RT\ln \left(\frac{p_B}{p^{\minuso}}\right) \end{align*} $$

Let us assume that for this general reaction the initial quantity of species A and B are $n_{A,0}$ and $n_{B,0}$. As the reaction proceeds, every mole of B created leads to a corresponding decrease in A, we can quantify this exchange of material by introducing the extent of the reaction, ($\xi$/ mol), such that: $$ n_A = n_{A,0} - \xi $$ $$ n_B = n_{B,0} + \xi $$ The only step remaining is to recognise that the partial pressure of the gas can be expressed as: $$ p_i = \chi_i p_{tot} = \frac{n_i}{n_{tot}} p_{tot}$$ where $\chi_i$ is the mole fraction of the species in the gas phase. Hence: $$ \begin{align*} G =~&(n_{A,0} - \xi) G^{\minuso}_{\mathrm{m},A} + (n_{B,0} + \xi) G^{\minuso}_{\mathrm{m},B} \\ &+ (n_{A,0} - \xi) RT\ln\left( \frac{(n_{A,0} - \xi)p_{tot}}{n_{tot}p^{\minuso}}\right) \\ &+ (n_{B,0} + \xi) RT\ln \left(\frac{(n_{B,0} + \xi)p_{tot}}{n_{tot}p^{\minuso}}\right) \end{align*} $$ Using this expression allows us to predict the variation in the system's Gibbs energy as a function of the extent of the reaction.

The interactive figure below plots the predicted Gibbs energy as a function of the extent of the reaction. Here, for the default parameters, $n_{A,0} =$ 1 mol, $n_{B,0} =$ 0 mol and the difference between the standard Gibbs energy of species A and B has been set to 2000 J mol^-1. At the start of the reaction, the Gibbs energy of the system is equal to 0 J, as only species A is present. However, although the energy of the 'product' species B is higher, the system can lower its total Gibbs energy by consuming some of A and converting it into species B. This decrease arises from the favourable entropy of mixing, having a mixture of both species A and B is more favourable from an entropic view. This plot goes through a minimum, which represents the lowest possible Gibbs energy for the system and corresponds to the equilibrium state. In the present example, this minimum occurs when the extent of the reaction is equal to approximately 0.31. For this system, as the reaction proceeds from its initial conditions, the Gibbs energy of the system reduces by roughly 900 J.

Using this interactive plot, explore for the reaction A ⇌ B:

How the position of the minimum varies as a function of the difference between the Gibbs energies of species A and B.
What happens to the difference between the initial and equilibrium energy (highlighted in red) if we double the initial concentration of species A?
What happens if species B is already present? Why can we now have a negative extent of reaction?

Gibbs Energy and Δ_rG vs. Reaction Extent (A ⇌ B)

G^°_A: 0 J mol^-1

G^°_B: 5 J mol^-1

n⁰_A: 1

n⁰_B: 0

ξ (for ΔG label): 0.00

Finding the Equilibrium

How can we find the position of this minimum in the Gibbs energy? This is a relatively straightforward task. We can take our expression for the Gibbs energy and differentiate it with respect to the extent of the reaction. At first, due to the logarithmic terms involving $ \xi $, it might appear messy, but many terms cancel, and we are left with: $$ \left. \frac{\partial G}{\partial \xi}\right|_{T,p} = G^{\minuso}_{\mathrm{m},B} - G^{\minuso}_{\mathrm{m},A} + RT\ln\left( \frac{p_B}{p_A} \right) $$

Given that the partial pressures are a function of the extent of the reaction, we can re-express this relation as: $$ \left. \frac{\partial G}{\partial \xi}\right|_{T,p} = G^{\minuso}_{\mathrm{m},B} - G^{\minuso}_{\mathrm{m},A} + RT\ln\left( \frac{(n_{B,0}+\xi)}{(n_{A,0}-\xi)} \right) $$ This expression is plotted in the second of the two interactive plots, showing the gradient of the Gibbs energy with respect to the extent of the reaction. As the extent of the reaction is measured in moles, then the gradient of the plot of Gibbs energy has units of J mol^-1.

This gradient is physically meaningful, it tells us the direction and magnitude of the thermodynamic driving force. First, at equilibrium; $$ \left. \frac{\partial G}{\partial \xi}\right|_{T,p} = 0 $$ Hence, recognising that the ratio of the two partial pressures is equal to the equilibrium constant $ K $, we obtain: $$\Delta G_m^{\minuso} = - RT \ln \left( \frac{p_B}{p_A} \right) = -RT \ln K $$ More broadly, this gradient is a measure of the driving force towards equilibrium: when negative, the system moves towards larger values of $\xi$; when positive, it moves towards smaller $\xi$. The gradient $\partial G/ \partial \xi |_{T,p}$ reflects the thermodynamic driving force. Historically, its negative was known as the chemical affinity A, a quantity introduced by de Donder^[2] to describe thermodynamic driving forces.

Returning to Chemical Potentials

But what controls this gradient? What is it made of, physically? It turns out that the driving force for the reaction can be decomposed into contributions from each chemical species. These contributions represent the individual chemical potentials of the species involved, indicating how much the Gibbs energy would change if a tiny amount of just one species were added. So while the extent of reaction tracks how the system composition changes, the chemical potential tells us what's happening for each individual species, and equilibrium is reached when those forces are balanced. The gradient $\partial G/ \partial \xi |_{T,p}$ for this simple reaction can be expressed as: $$ \left. \frac{\partial G}{ \partial \xi} \right|_{T,p} = \mu_B - \mu_A $$ where: $$\mu_A = \left. \frac{\partial G}{\partial n_A}\right|_{T,p}$$ $$\mu_B = \left. \frac{\partial G}{\partial n_B}\right|_{T,p}$$

At first glance, this might seem like a step back. We've already got an expression for the gradient of G, so why define these chemical potentials? But in fact, the chemical potential is more than a convenient variable, it turns out to be the natural thermodynamic potential for composition. Once defined, it becomes a tool that can be applied far beyond chemical reactions: in phase equilibrium, diffusion, and electrochemistry.

In the first example, we considered how the chemical potential could be used to describe the physical partitioning of a species between two boxes. That same logic applies here to the case of a chemical reaction, except instead of spatial flow, we consider “reaction flow” between species. Just as matter moves from high to low chemical potential, a chemical reaction proceeds in the direction that lowers the Gibbs energy, which is determined by the difference in chemical potentials.

So by using the extent of reaction as a coordinate, we encode the behaviour of both species A and B. More generally: $$ \left. \frac{\partial G}{\partial \xi}\right|_{T,p} = \sum_i \nu_i \mu_i $$ where $ \nu_i $ is the stoichiometric coefficient. The differences in the chemical potentials of the species tell us about the thermodynamic driving force for moving towards equilibrium and the values are related to the gradient of the Gibbs energy for a system.

A Notational Subtlety, $ \Delta_r G $

We've now arrived at an important point: we understand how the Gibbs energy of a reacting system changes as the composition evolves, and we've seen that the slope indicates whether the reaction wants to proceed. Returning to the gradient of the Gibbs energy, for our simple reaction, we have A ⇌ B: $$ \left. \frac{\partial G}{\partial \xi}\right|_{T,p} = \Delta G^{\minuso}_{\mathrm{m},B} - \Delta G^{\minuso}_{\mathrm{m},A} + RT\ln\left( \frac{p_B}{p_A} \right) $$ The term on the right-hand side is just the difference between the molar Gibbs energies of the two species A and B; this difference in the molar Gibbs energies is often denoted as the reaction Gibbs energy. $$ \begin{align*} \Delta_r G &= \Delta_r G^{\minuso} + RT\ln\left( \frac{p_B}{p_A} \right) \\ &= \Delta G^{\minuso}_{\mathrm{m},B} - \Delta G^{\minuso}_{\mathrm{m},A} + RT\ln\left( \frac{p_B}{p_A} \right) \end{align*} $$ This is where confusion can arise, the gradient of the plot of the Gibbs energy versus the extent of the reaction just happens to be equal to the difference in the molar Gibbs energies. Here, $$ \left. \frac{\partial G}{\partial \xi}\right|_{T,p} = \Delta_r G$$ So $\Delta_r G$, is actually a measure of the gradient,^[3] it is describing the driving force for a change in a chemical reaction. The problem is that most of the time, when we have a $\Delta $, we are representing a large change, like: $$ \Delta G = G_{\mathrm{final}} - G_{\mathrm{initial}} $$ but this isn't the case here. In some ways this is a notational failing or perhaps just a notational ambiguity. However, $\Delta_r G$ is a difference between two molar Gibbs energies and we use it to describe a gradient. The use of the $\Delta$ symbol can lead us to misinterpret the value as a large macroscopic change.

On the interactive plot, I have added a slider where you can look at the value of $\Delta_r G$ at different points through the course of the reaction and we should now recognise why, $$ \Delta_r G \neq G_{\mathrm{equilibrium}} - G_{\mathrm{initial}} $$

As a very final point, this issue in terms of misinterpreting $ \Delta_r G $, also comes up in electrochemistry, when we have the expression: $$ \Delta_r G = - nFE $$ Just as $\Delta_r G$ provides a measure of the chemical driving force, the electrode potential E reflects the electrical driving force in an electrochemical cell. The relation $\Delta_r G = -nFE$ again links Gibbs energy to a measurable driving force, with the same caution that this is not a simple energy difference, but a gradient-driven relation.

Footnotes

[1] Just as a reminder temperature is controlled by the total amount of energy in a substance. Imagine if we had a 30^oC 1 kg block and then replicated it perfectly, it too would have exactly the same energy and so it would be at the same temperature. Adding this new block to the system would increase the total energy of the system but the temperature would remain the same.

[2] Thermodynamic Theory of Affinity, : A Book of Principles. de Donder, Théophile (1936). This is the English translation of his work, it is not a light read. Just to note, although it isn't often used affinity is still defined in the IUPAC green book.

[3] As expressed in the IUPAC Green handbook, $\Delta_r$ can be interpreted as operator symbol $ \Delta_r \overset{\mathrm{def}}{=} \partial / \partial \xi$.